You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 50.0 m away, making a 2.00degrees angle with the ground.How fast was the arrow shot?
To calculate the initial speed of the arrow, you can use the equation of projectile motion. In this case, since the arrow is shot parallel to the ground, the initial vertical velocity is zero. The equation relating the horizontal distance, preliminary angle, and initial speed of the arrow is as follows:
d = (v^2 * sin(2θ)) / g,
where,
d is the horizontal distance traveled by the arrow (50.0 m),
θ is the preliminary angle with the ground (2.00 degrees),
v is the initial speed of the arrow, and
g is the acceleration due to gravity (9.8 m/s^2).
First, convert the angle from degrees to radians. Since 180 degrees = π radians, the preliminary angle in radians (θr) can be calculated as follows:
θr = (2.00 degrees) * (π / 180 degrees) = 0.0349066 radians.
Next, rearrange the equation to solve for the initial speed (v):
v = sqrt((d * g) / sin(2θ)).
Substitute the given values into the equation:
v = sqrt((50.0 m * 9.8 m/s^2) / sin(2 * 0.0349066 radians)).
Using a calculator, evaluate sin(2 * 0.0349066 radians), which results in approximately 0.0697565.
v = sqrt((50.0 m * 9.8 m/s^2) / 0.0697565).
v = sqrt(705.88 m^2/s^2) ≈ 26.56 m/s.
Therefore, the arrow was shot with an initial speed of approximately 26.56 m/s.